Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.821 M and [Ni2 ] = 0.0200 M. Standard reduction potentials can be found here.

Zn(s) + Ni+2(aq)=Zn+2(aq) + Ni(s)

Answer #1

we have use the equation

E cell = E^{0} cell - 0.059/n logQ

first find out the E^{0} cell using standard reduction
potential values

Ni^2+ + 2e- --> Ni(s) E0: -0.28V

Zn^2+ + 2e- --> Zn(s) E0: -0.763V

Zn+ Ni^2+ --> Zn^2+ Ni .. E0: -0.28 -(-0.76) = 0.48V

this is the E^{0} of the cell

Q = [Zn2+] / [Ni2+] = 0.821 M / 0.02 M = 41.05

log Q = log(41.05) = 1.61

n= no of electrons involved here that is 2

plug in all the values in the above equation

E cell = 0.48 V - (0.059 / 2 ) (1.61)

E cell = 0.48 V - (0.0475 V)

= 0.43V

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