Question

Consider two hypothetical metals, X and Y, which can exist as metal ions X2+ and Y3+, respectively. For these two metals, the standard reduction potentials are X 2+ + 2 e − → X E 0 = −0.550 V Y 3+ + 3 e − → Y E 0 = 2.100 V Calculate Ecell for the reaction 3 X(s) + 2 Y3+(0.250 M) → 3 X2+(0.0500 M) + 2 Y(s). Answer in units of volts.

Answer 1

In the cell X is getting oxidised and Y is reduced

Eo cell = Eo cathode - Eo anode
= 2.100 V - (-0.550 V)
= 2.650 V

Number of electron being transferred in balanced reaction is 6
So, n = 6

use:
E = Eo - (2.303*RT/nF) log {[X2+]^3/[Y3+]^2}

Here:
2.303*R*T/F
= 2.303*8.314*298.0/96500
= 0.0591

So, above expression becomes:
E = Eo - (0.0591/n) log {[X2+]^3/[Y3+]^2}
E = 2.650 - (0.0591/6) log (0.05^3/0.25^2)
E = 2.650-(-2.66*10^-2)
E = 2.677 V
Answer: 2.677 V