What is the standard emf of a galvanic cell made of a Sn electrode in a 1.0 M Sn(NO_3)_2Sn(NO3)2 solution and a Cu electrode in a 1.0 M Cu(NO_3)_2Cu(NO3)2 solution at 25°C?

1 Answer
Dec 9, 2015

E_(cell)^@=+0.47"V"Ecell=+0.47V

Explanation:

You need to look up standard electrode potentials and list them -ve to +ve:

" "E^@("V") E(V)
stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(red)(larr))×××××××××−−−−−−−−−−−

Sn_((aq))^(2+)+2erightleftharpoonsSn_((s))" "-0.13Sn2+(aq)+2eSn(s) 0.13

Cu_((aq))^(2+)+2erightleftharpoonsCu_((s))" "+0.34Cu2+(aq)+2eCu(s) +0.34
stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(blue)(rarr))×××××××××−−−−−−−−−−−

You can see that the more +ve 1/2 cell will take in the electrons so the 1/2 cell reactions proceed in the direction indicated by the arrows.

Emf is an experimentally measured quantity and must always have a +ve value.

If you try to measure the emf of a cell and get a -ve reading, it means you have connected the voltmeter to the wrong terminals.

So to get the emf of the cell, always subtract the least positive potential from the most positive potential rArr

E_(cell)^@=+0.34-(-0.13)=+0.47"V"Ecell=+0.34(0.13)=+0.47V

I have adopted the convention which is used in the UK.

I understand other conventions would reverse the sign of the Sn^(2+)"/"SnSn2+/Sn 1/2 cell then add.

Other conventions I have seen write the 1/2 cells in the other direction and reverse the voltage.