I have an electrochemical cell with following overall reaction: $$\ce{5 Sn^4+ + 2 Mn^2+ + 8 H2O <=> 5 Sn^2+ + 2 MnO4- + 16 H+}$$

I have defined the half reactions as:

$$ \begin{align} \ce{Sn^2+ &-> Sn^4+ + 2 e-} \\ \ce{MnO4- + 8 H+ + 5 e- & -> Mn^2+ + 4 H2O} \end{align} $$

I have trouble calculating the biological standard electrochemical potential. This is what I've tried:

The standard potential can be found from tables:

$$E^\circ = \pu{1.51 V} - \pu{0.15 V} = \pu{1.36 V}$$

Then I use Nernst equation in which all species except $\ce{H+}$ are in their standard states, so their activities are all equal to $1:$

$$ \begin{align} E &= E^\circ - \frac{RT}{vF}\ln(10)\cdot \mathrm{pH} \\ &= \pu{1.36 V} - \frac{\pu{8.314 J mol-1 K-1} × \pu{298.15 K}}{10 × \pu{96485 C mol-1}}\ln(10) × 7 \\ &= \pu{1.31 V}, \end{align} $$

which is wrong. Is there something wrong with the calculations or the understanding?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Commented May 22, 2020 at 20:04

1 Answer 1


Unfortunately, the OP's Nernst equation is incorrect for several reasons. The balanced net redox equation is

$$\ce{5 Sn^{4+} (aq) + 2 Mn^{2+} (aq) + 8 H2O <=> 5 Sn^{2+} (aq) + 2 MnO4^- (aq) + 16 H+ (aq) \tag 1}$$ and n = 10 electrons for the reaction as written. The reaction quotient, Q, is

$$\ce{Q = \frac{[Sn^{2+}]^5 [MnO4^-]^2 [H+]^16}{[Sn^{4+}]^5 [Mn^{2+}]^2} \tag 2}$$

and the Nernst equation is

$$\displaystyle{E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln(Q) = E_{cell}^\circ - \frac{RTln(10)}{nF} \log(Q) \tag 3} $$

The OP's question states that "all species except $\ce{H+}$ are in their standard states, so their activities are all equal to 1". Hence, in this very contrived example, Q reduces dramatically:

$$\displaystyle{Q = [H^+]^{16} \tag 4}$$

The OP notes that $\displaystyle{E_{cell}^\circ }$ = 1.36 V (from tables). With T = 298.15 K, $R = \pu{8.31446261815324 J mol-1 K-1}$ and $F = \pu{96485.3321233100184 C mol-1}$, then

$$\ce{\displaystyle{ \frac{RTln(10)}{nF} \approx \frac{0.05916 \ V}{n} } \tag 5}$$

So the Nernst equation is simply

$$\displaystyle{E_{cell} \approx 1.36 \ V - \frac{0.05916 \ V}{n} \log(Q) = 1.36 \ V - \frac{0.05916 \ V}{10} \log([H^+]^{16}) \tag 6} $$

Hence, using $\ce{pH = -log[H^+] = +log(1/[H^+])}$, the Nernst equation can be expressed as

$$\displaystyle{E_{cell} \approx 1.36 \ V - 16 \times \frac{0.05916 \ V}{10} \log[H^+] = 1.36 \ V + 16 \times \frac{0.05916 \ V}{10} pH \tag 7} $$

If pH = 7, then $\ce{E_{cell} \approx 2.02 \ V}.$

It is better to use expression (3) in order to avoid getting too tricky. This is where the OP went wrong: forgetting the "16" exponent on the hydrogen ion concentration and also missing the fact that the negative sign before the pre-log correction term must be subsumed into the pH definition. In a realistic example, it will not happen that everything is in standard state except the hydrogen ion.

What about dividing the net ionic equation, i.e., equation (1), by 16? Then n = 10/16 = 5/8 and all the exponents in the Q expression also get divided by 16. This is no problem if everything is in standard state except the hydrogen ion. Otherwise, it is just facilitating making a silly mistake.

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    – Ed V
    Commented May 26, 2020 at 1:33

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