Write Nernst equation to calculate cell potential of above cell
Mg (s) | Mg2+ (aq) || Ag+ (aq) | Ag (s)
Dear student
Please find the solution to the asked query:
Reactions Mg is oxidised to Mg2 +
Mg(s) => Mg2 +(aq) + 2e-
Ag+ is reduced to Ag
[Ag+ (aq) + e- => Ag(s)] x 2
Nernst equation E = Eo - RT x 2.303/2F log[Mg2+]/[Ag+]2
where,
E0 = cell potential under standard conditions
R = gas constant
T = temperature (K)
n = number of moles of electrons exchanged in the electrochemical reaction (mol) ans equals 2 in this case
F = Faraday's constant
Hope this information will clear your doubts regarding the topic..If you have any other doubts please ask here on the forum and our experts will try to solve them as soon as possible,
Regards
Please find the solution to the asked query:
Reactions Mg is oxidised to Mg2 +
Mg(s) => Mg2 +(aq) + 2e-
Ag+ is reduced to Ag
[Ag+ (aq) + e- => Ag(s)] x 2
Nernst equation E = Eo - RT x 2.303/2F log[Mg2+]/[Ag+]2
where,
E0 = cell potential under standard conditions
R = gas constant
T = temperature (K)
n = number of moles of electrons exchanged in the electrochemical reaction (mol) ans equals 2 in this case
F = Faraday's constant
Hope this information will clear your doubts regarding the topic..If you have any other doubts please ask here on the forum and our experts will try to solve them as soon as possible,
Regards