Refer to galvanic cell. 0.010 M Zn 2+ in left cell and 1.5 M Ag + in right cell.
ID: 836623 • Letter: R
Question
Refer to galvanic cell. 0.010 M Zn2+ in left cell and 1.5 M Ag+ in right cell.
Standard reduction potentials are as follows:
Zn2+ + 2e- --> Zn E= -0.76 V
Ag+ + e- --> Ag E= 0.80 V
1) When current is allowed to flow, which species is oxidized?
a) Zn2+ b) Ag+ c) Ag d) Zn
2) What is the value of Q the reaction quotient, for the above cell reaction?
a) 225 b) 1 c) 4.44x10-3 d) 150
3) Based on the setup of the voltaic cell in the above diagram, what is the actual cell potential E for the above cell at 25C
a) 1.82 V b) 1.56V c) 1.64V d) 0.04V
Explanation / Answer
1. d) Zn is oxidised to Zn2+
2. The overall reaction is
Zn + 2Ag+ --------------- Zn2+ + 2Ag
Therefore, Reaction Quotient = Q = [Zn2+] / [Ag+]2 = (0.010) / (1.5)2 = b) 4.44 x 10-3
2. Eo = Ecathode - Eanode = 0.80 + 0.76 = 1.56V
Ecell = Eo - {(R*T)/(n * F)} * ln Q
= 1.56 - (0.059/2) * 4.44 * 10-3 = 1.5598V = b) 1.56V(appx.)