Question

Consider the galvanic cell described by (N and M are metals): N(s)|N2+(aq)||M+(aq)|M(s) If Eocathode = 0.047...

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

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

If Eocathode = 0.047 V and Eoanode = 1.129 V, and [N2+(aq)] = 0.281 M and [M+(aq)] = 0.775 M, what is Ecell, using the Nernst equation? ____ V

Provide your response to two digits after the decimal.

Homework Answers

Answer #1

Eocell = Eo cathode - Eo anode

= 0.047 V - 1.129 V

= -1.082 V

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 = -1.082 - (0.0591/2) log (0.281^1/0.775^2)

E = -1.082-(-9.753*10^-3)

E = -1.072 V

Answer: -1.07 V

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