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Question:

Balance the following reaction: $$\ce{CuS + SO4^{2-} -> CuO + SO2}$$

BalanceBelow, the following reaction:number mentioned in the parenthesis is the oxidation number of sulfur.
$\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts: Number mentioned in the brackets oxidation number of S

II have found out that the oxidation half-reactions and reduction half-reactions are as follows:

$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction\begin{align} \ce{CuS^{(-2)} &-> ^{(4)}SO2}&& \text{Oxidation half reaction}\\ \ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction half reaction} \end{align}

Problem:

In this, what about $\ce{CuO}$ which in the product side.? I have also learned that we need to balance the spectator ionions before charges if the spectator ion is other than oxygen and hydrogen.

Here, there is no $\ce{Cu}$ in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.

Question:

Balance the following reaction: $\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts: Number mentioned in the brackets oxidation number of S

I have found out that the oxidation half-reactions and reduction half-reactions are as follows:

$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction

Problem:

In this what about $\ce{CuO}$ which in the product side. I have also learned that we need to balance the spectator ion before charges if the spectator ion is other than oxygen and hydrogen.

Here there is no $\ce{Cu}$ in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.

Balance the following reaction: $$\ce{CuS + SO4^{2-} -> CuO + SO2}$$

Below, the number mentioned in the parenthesis is the oxidation number of sulfur.
I have found out that the oxidation half-reactions and reduction half-reactions are as follows:

\begin{align} \ce{CuS^{(-2)} &-> ^{(4)}SO2}&& \text{Oxidation half reaction}\\ \ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction half reaction} \end{align}

Problem:

In this, what about $\ce{CuO}$ in the product side? I have also learned that we need to balance the spectator ions before charges if the spectator ion is other than oxygen and hydrogen.

Here, there is no $\ce{Cu}$ in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.

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jonsca
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Balancing redox reaction by What do the half-reactions need to be for balancing a redox reaction methodin an acidic medium?

Question:

Balance the following reaction: $\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts: Number mentioned in the brackets oxidation number of S

I have found out that the oxidation half reaction-reactions and reduction half reaction which-reactions are as follows:

$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction

Problem:

In this what about $\ce{CuO}$ which in the product side. I have also learned that we need to balance the spectator ion before charges if the spectator ion is orher thenother than oxygen and hydrogen.

Here there is no $\ce{Cu}$ in reduction half reaction. So what i, I think there is something is wrong in my reduction half reaction-reaction.

PLEASE HELP

Balancing redox reaction by half reaction method

Question:

Balance the following reaction: $\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts: Number mentioned in the brackets oxidation number of S

I have found out oxidation half reaction and reduction half reaction which are as follows

$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction

Problem:

In this what about $\ce{CuO}$ which in the product side. I have also learned that we need to balance the spectator ion before charges if spectator ion is orher then oxygen and hydrogen.

Here there is no $\ce{Cu}$ in reduction half reaction. So what i think there is something is wrong in my reduction half reaction.

PLEASE HELP

What do the half-reactions need to be for balancing a redox reaction in an acidic medium?

Question:

Balance the following reaction: $\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts: Number mentioned in the brackets oxidation number of S

I have found out that the oxidation half-reactions and reduction half-reactions are as follows:

$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction

Problem:

In this what about $\ce{CuO}$ which in the product side. I have also learned that we need to balance the spectator ion before charges if the spectator ion is other than oxygen and hydrogen.

Here there is no $\ce{Cu}$ in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.

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Freddy
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Question:

Balance the following reaction: $\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts: Number mentioned in the brackets oxidation number of S

I have found out oxidation half reaction and reduction half reaction which are as follows

$\ce{CuS-> SO2}$$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{SO4^{2-} -> SO2}$$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction

Problem:

In this what about $\ce{CuO}$ which in the product side. I have also learned that we need to balance the spectator ion before charges if spectator ion is orher then oxygen and hydrogen.

Here there is no $\ce{Cu}$ in reduction half reaction. So what i think there is something is wrong in my reduction half reaction.

PLEASE HELP

Question:

Balance the following reaction: $\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts:

I have found out oxidation half reaction and reduction half reaction which are as follows

$\ce{CuS-> SO2}$ Oxidation half reaction

$\ce{SO4^{2-} -> SO2}$ Reduction half reaction

Problem:

In this what about $\ce{CuO}$ which in the product side. I have also learned that we need to balance the spectator ion before charges if spectator ion is orher then oxygen and hydrogen.

Here there is no $\ce{Cu}$ in reduction half reaction. So what i think there is something is wrong in my reduction half reaction.

PLEASE HELP

Question:

Balance the following reaction: $\ce{CuS + SO4^{2-} -> CuO + SO2}$

My Efforts: Number mentioned in the brackets oxidation number of S

I have found out oxidation half reaction and reduction half reaction which are as follows

$\ce{CuS^{(-2)}-> ^{(4)}SO2}$ Oxidation half reaction

$\ce{^{(6)}SO4^{2-} -> ^{(4)}SO2}$ Reduction half reaction

Problem:

In this what about $\ce{CuO}$ which in the product side. I have also learned that we need to balance the spectator ion before charges if spectator ion is orher then oxygen and hydrogen.

Here there is no $\ce{Cu}$ in reduction half reaction. So what i think there is something is wrong in my reduction half reaction.

PLEASE HELP

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Freddy
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