Hi
question is which of the following changes will raise the voltage in the galvanic cell
ZnZn^2+\Cu^2+Cu
E = 1.1 V
A. raising concentrations of both [Zn^2+] and [Cu^2+] at the same amount.
B. [Zn^2+] remains the same and raising [Cu^2+] concentration.
C. [Cu^2+] remains the same and raising [Zn^2+] concentration.
D. lowering both [Zn^2+] and [Cu^2+] concentrations at the same amount.
I am pretty sure it’s C because of Nernst equation the part of log[Ox]/[Red] , raising the Ox part will raise the voltage, therefore it’s Zn (the atom being oxidize).
Is that correct?
Thank you.
2
✅ Answers
? Favorite Answer
oxidtion half :-
Zn ——–> Zn2+ + 2e-
reduction half :-
Cu2+ + 2e- ——> Cu
complete equation :-
Zn + Cu2+ ———–> Zn2+ + Cu
nernst equation …
E(cell) = Eo(cell) – RT/nF ln [Zn2+]/[Cu2+]
now as 2e- are involved in electrode reaction so n = 2
E(cell) = Eo(cell) – RT/2F ln [Zn2+]/[Cu2+]
now here Eo(cell) is constant = 1.1 V ….
RT/2F is also constant …
lets just say that temperature is 298 K
then RT/ 2F = 8.314 X 298 / 2 X 965 = 0.0129
lets see what happen when you increase the conc.of Zn2+ keeping the conc.of Cu2+ constant …..say initially [Cu2+] = 1M and [Zn2+] = 2 M
then E(cell) = 1.1 – 0.0129 X ln 2/1 = 1.1 – 0.0129 X ln 2 = 1.1 – 8.942 X 10^-3 = 1.091 V
now say you increase the conc. of Zn2+ to 4 M keeping Cu2+ at 1 M
then E(cell) = 1.1 – 0.0129 X ln 4/1 = 1.1 – 0.0129 X ln 4 = 1.1 – 0.0179 = 1.082 V
so E(cell) has decreased when you raise the conc.of Zn2+ keeping the Cu2+ conc. constant ..
———————————————————————–
now say Zn2+ = 1 M and Cu2+ = 2 M
then E(cell) = 1.1 – 0.0129 X ln 1/2 = 1.1 – 0.0129 X ln0.5 = 1.1 – -8.942 X 10^-3 = 1.1 + 8.942 X 10^-3 = 1.109 V
and now again Zn2+ = 1 M and Cu2+ = 4 M
then E(cell) = 1.1 – 0.0129 X ln 1/4 = 1.1 – 0.0129 X ln0.25 = 1.1 – -0.018 = 1.1 + 0.018 = 1.118 V
so E(cell) increases when you increase the conc.of Cu2+ keeping conc.of Zn2+ constant
so answer is B
C