Q13 of 26 Page 1

Calculate emf of following cell at 298 K

Sn|Sn2+(0.001M)||H+(0.01M)||H2(g)(1bar)|Pt (s)


E°(Sn2+/Sn) = -0.14V


E ° (H+/H2) = 0.00V


From E° values, we can conclude that Sn will undergo oxidation due to lower reduction electrode potential and hydrogen undergo oxidation.

E° cell= E°cathode-E°anode


= 0.00 – (-0.14) = 0.14 V


Ecell = E°cell – log


Ecell = 0.14 –log


Ecell = 0.14 –log


= 0.14- 0.0295[-3-(-4) ]


0.14 – 0.0295(1)


=0.1105 V


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