Question

Calculate the cell emf for the following reaction at 25°C: 2Ag+(0.010 M) + H2(1 atm) ? 2Ag(s) + 2H+(pH = 8.00)

Answer 1

First, some tabulated data:
Ag+ (aq) + e− ==> Ag(s), Eº = 0.80
2 H+ (aq) + 2 e− ==> H2(g), Eº = 0.00

Since your cell isn't in standard (all solutions 1 M, all gases 1 atm), you must use Nernst's equation to account for this difference. It states that:
E = Eº - (RT/nF) ln([Red]/[Ox]),
where R and F are the ideal gas' and Faraday's, constants respectively, n the ammount of electrons involved and [Red] and [Ox] the concentrations of the oxidated and reduced compounds involved, respectively. Since no temperature is stated, we'll supose the standard 25ºC, and rewrite the equation as:
E = Eº - (0.0591/n) log([Red]/[Ox]).
Solids and the solvent have an activity of 1, and thus can be left out of this equation. In the case of gases, their partial pressure is used instead of their concentration.

So, for the silver's half-reaction:
Ag+ (aq) + e− ==> Ag(s), Eº = 0.80
And E = Eº - (0.0591/n) log([Red]/[Ox]) becomes:
E = 0.80 - (0.0591/1) log(1/[Ag+])
E = 0.80 + 0.0591 log([Ag+])
For [Ag+] = 0.010 M,
E = 0.80 + 0.0591 log(0.010)
E = 0.80 - 2*0.0591
E = 0.6818

And for hydrogen's:
2 H+ (aq) + 2 e− ==> H2(g), Eº = 0.00
And E = Eº - (0.0591/n) log([Red]/[Ox]) becomes:
E = - (0.0591/2) log((H2)/[H+]^2)
E = - 0.02955 log((H2)/[H+]^2)
E = - 0.02955 log((H2)) + 0.0591 log([H+])
E = - 0.02955 log((H2)) - 0.0591 pH
For (H2) = 1 atm and pH = 8.0,
E = - 0.02955 log(1) - 0.0591*8.0
E = - 0.4728

So, the cell's E is:
Ecell = Ered - Eox
Ecell = 0.6818 + 0.4728
Ecell = 1.1546 V