Question

Calculate Ecell for the reaction below when [Zn2+] = 1.00 M, [H+] = 1.00 x 10–6 M, and P H2 = 1.00 atm and a temperature of 25°C.

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

(a) +0.41 V (b) +0.053 V (c) 0.64 V (d) +1.12 V (e) +0.76 V

The correct answer is A, however, I dont know how to calculate this on my own. Can someone explain and show the steps in depth please? Thank you!

Answer 1

Eo(Zn2+/Zn) = -0.76 V
Eo(H+/H2) = 0 V

reaction at cathode :-

2H+ + 2e- -------> H2 .....Eo(cathode) = 0 V

and reaction at anode :-

Zn ----------> Zn2+ + 2e- .....Eo(anode) = -0.76 V

so Eo(cell) = Eo(cathode) - Eo(anode) = 0 - -0.76 = 0 + 0.76 = 0.76 V

now if the concentration of Zn2+ , H+ had been 1 M and PH2 were 1 atm ...E(cell) would have been equal to Eo(Cell )

but since it is not ...so using nernst equation ...

E(cell) = Eo(cell) - 2.303RT/nF log [Zn2+] PH2 / [H+]^2

where R = 8.314 J/K/mole
T = 298 K
n = no.of electrons involved in electrode reaction
F = 96500 C/mole
[Zn2+] = 1 M
[H+] = 10^-6 M
PH2 = 1 atm

E(cell) = 0.76 - 2.303 X 8.314 X 298 / ( 2 X 96500) log 1 X 1 / ( 10^-6)^2

E(cell) = 0.76 - 0.0296 X log 10^12 = 0.76 - 0.355 = 0.405 V