Question

A voltaic cell is constructed that uses the following reaction and operates at 298 K.

Zn(s) + Ni2+(aq) Zn2+(aq) + Ni(s)

(a) What is the emf of this cell under standard conditions? V

(b) What is the emf of this cell when [Ni2+] = 2.60 M and [Zn2+] = 0.194 M? V

(c) What is the emf of the cell when [Ni2+] = 0.232 M and [Zn2+] = 0.979 M? V

Answer 1

First calculate Eo for the cell.

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

Ni2+(aq) + 2e- ==> Ni(s) . . . Eo = -0.23 V

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Zn(s) + Ni2+(aq) ==> Zn2+(aq) + Ni(s) . .Eo cell = +0.53 V

Given that [Ni2+] = 2.60 M and [Zn2+] = 0.194 M

Then use the Nernst equation:

E cell = Eo cell - 0.059/n log Q

= 0.53 - 0.059/2 log ([Zn2+]/[Ni2+])

= 0.53 - 0.0295 log (0.194/2.60)

= 0.53 - 0.0295 log0.0746

=0.53 - 0.0295 *(-1.13)

=0.53 +0.033

= 0.563 V

For part C)

[Ni2+] = 0.232 M and [Zn2+] = 0.979 M

Then use the Nernst equation:

E cell = Eo cell - 0.059/n log Q

= 0.53 - 0.059/2 log ([Zn2+]/[Ni2+])

= 0.53 - 0.0295 log (0.979/2.32)

= 0.53 - 0.0295 log 0.422

=0.53 - 0.0295 *(-0.374)

=0.53 +0.011

= 0.541 V