1
$\begingroup$

I was presented with the following problem and I am finding it difficult to present the half equations for the answers.

Redox titration

I managed to produce the half-equation for the Iodine redox.

$\ce{2I- -> I2 + e-}$

I then proceeded to the $\ce{CrO4^{2-}}$ equation.

$\ce{CrO4^{2-} + 3e- -> Cr3+}$

This is because the oxidation number of the Cr in $\ce{CrO4^{2-}}$ is $+6$ and it is reduced to $+3$

However, the answer is given as:

$\ce{CrO4^{2-} + 4H2O + 3e- → Cr3+ + 8OH-}$

I am struggling to get from the equation I made to the one presented in the answer for the $\ce{CrO4^{2-}}$ redox.

Can someone explain why there are $\ce{4H2O}$ on the left hand side and $\ce{8OH-}$ on the left hand side of the equation?

Please help! Many thanks.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.