Question

Consider the galvanic cell described by (N and M are metals):

N(s)|N2+(aq)||M+(aq)|M(s)

If Eocathode = 0.213 V and Eoanode = 1.959 V, and [N2+(aq)] = 0.63 M and [M+(aq)] = 0.044 M, what is Ecell, using the Nernst equation? ____ V

Provide your response to two digits after the decimal.

Answer 1

Eocell = Eoathode - Eoanode

= 0.213 - 1.959

= -1.746

the reaction is:

2 M+ (aq) + N(s) —> 2 M (s) + N2+ (aq)

Number of electron being transferred in balanced reaction is 2

So, n = 2

we have below equation to be used:

E = Eo - (2.303*RT/nF) log {[N2+]^1/[M+]^2}

Here:

2.303*R*T/n

= 2.303*8.314*298.0/F

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[N2+]^1/[M+]^2}

E = -1.746 - (0.0591/2) log (0.63^1/0.044^2)

E = -1.746-(7.428*10^-2)

E = -1.82 V

Answer: -1.82 V