# How do I calculate the cell voltage using the following half-reactions and solubility products?

The concentrations of $\ce{NaCl}$ and $\ce{KI}$ are at $\pu{0.120 M}$ and $\pu{0.090 M}$, respectively, in the following electrochemical cell:

$$\ce{Cu_{(s)}|CuI_{(s)}|I^{–}_{(aq)}||Cl^{–}_{(aq)}| AgCl_{(s)}|Ag_{(s)}}$$

How do I calculate the cell voltage using the following half-reactions:

\begin{align} \ce{Cu+ + e- &<=> Cu(s) & E^\circ &= \pu{0.518 V}} \\ \ce{Ag+ + e- &<=> Ag(s) & E^\circ &= \pu{0.7993 V}} \end{align}

The solubility products $(K_{sp})$ for $\ce{AgCl}$ and $\ce{CuI}$ are $\pu{1.8e–10}$ and $\pu{1.0e-12}$, respectively.

I think you need the Nernst equation. This is something we were just introduced to today in my class, so I’m not so sure about the notation you’re using in your question at the top, but you should be able to correct my answer to fit your needs if you understand the equation I’m about to implement.

$$E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{zF}\ln Q$$

Where $E^\circ_\text{cell}$ = what the $E_\text{cell}$ normally is, under STP with the solutes at a concentration of $1\ \mathrm{mol\ dm^{−3}}$.

$$E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}$$

Since, the reduction potential of copper is lower, it will be the reducing agent in this reaction.

$$E^\circ_\text{cell} = -0.7993\ \mathrm{V} + 0.518\ \mathrm{V}$$

$$E^\circ_\text{cell} = -0.281\ \mathrm{V}$$

Where $R$ is the familiar gas law constant:

$$R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$$

$T$ is the absolute temperature (it’s always assumed to be $298.15\ \mathrm{K}$):

$$T = 298.15\ \mathrm{K}$$

$z$ is the number of moles of electrons transferred in the half-reaction:

$$z = 1$$

And $F$ is Faraday’s constant, which is measured in coulombs per mole:

$$F = 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}$$

And $Q$ is the also familiar reaction quotient, but as I discovered, it takes some work to calculate, and you have to use the quadratic equation.

$$Q=\frac{[\ce{red}]}{[\ce{ox}]}$$ $$Q=\frac{[\ce{Cu+}]}{[\ce{Ag+}]}$$

$$K_{\text{sp},\ce{CuI}}=10^{-12}=[\ce{Cu+}][\ce{I-}]$$ $$1.0 × 10 ^{-12} = [X][0.090 + X]$$ Do the quadratic formula $$X = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$ $$[\ce{Cu^+}] = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$ $$K_{\text{sp},\ce{AgCl}} = 1.8 \times 10^{-10}\ \mathrm{mol\ l^{-1}} = [\ce{Ag+}][\ce{Cl-}]$$ $$1.8 \times 10^{-10} = [X][0.120 + X]$$ Do the quadratic formula $$X = 1.50 \times 10 ^{-9}\ \mathrm{mol\ l^{-1}}$$ $$[\ce{Ag+}]=1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}$$

$$Q=\frac{1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}}{1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}}$$

$$Q=0.00740$$

So, plugging this all into the Nernst equation, the answer should be:

$$E_\text{cell}=-0.281\ \mathrm{V}-\frac{8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} \times 298.15\ \mathrm{K}}{1 \times 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}}\ln(0.00740)$$

$$E_\text{cell}=-0.155\ \mathrm{V}$$

I should add that since $R$ is a constant, $F$ is a constant, and $T$ is usually something like $298.15\ \mathrm{K}$ or just not measured in elementary chemistry, the equation can just be

$$E_\text{cell}=E^\circ_\text{cell}-\frac{0.0257}{z}\ln Q$$

Which will give the same answer:

$$E_\text{cell}=-0.281\ \mathrm{V}-0.0257\ \mathrm{V}\ln(0.00740)$$ $$E_\text{cell}=-0.155\ \mathrm{V}$$

• Generally you can neglect X and avoid the quadratic formula. .1 plus a very small number is still .1. – LordStryker May 16 '15 at 16:21