Answer to Question #218336 in Civil and Environmental Engineering for syra

Question

A 1 M solution of Cu(NO3)2 is placed in a beaker with a strip of Cu metal. A 1M solution of SnSO4 is placed in a second beaker with a strip of Sn metal. A salt bridge connects two beakers and wires to a voltmeter link the two metal electrodes. [ Hint: Charge of nitrate anion is (-1); Charge of Sulfate ion is (-2)]

What is the emf generated by the cell under standard conditions?

Answer 1

First, determine the reduction reaction for both copper and tin.

$C u^{2+} ( a q ) + 2 e^ − → C u^ o ( s )\\ S n^{ 2 +} ( a q ) + 2 e^ − → S n^ o$

The nitrate and sulfate ions are spectator ions so we can neglect them. One of these reactions goes as written and the other one must be written as oxidation.

The standard reduction potentials, E^o, of the two reactions are +0.34 V and -0.14 V. In order for the reaction to be spontaneous, Gibb's free energy of the reaction must be negative. Since the Gibb's free energy is a negative value when the electrical potential is positive, the copper must be reduced and the tin oxidized.

The second equation proceeds in reverse fashion.

$Sn^o (s) \rightarrow Sn^{2+} (aq)+ 2\: e^-$

The electrochemical reaction is

$C u^{ 2 +} ( a q ) + S n^ o ( s ) → C u^ o ( s ) + S n^{ 2 +} ( a q )$

Two electrons appeared on each side of the equation so they were eliminated. If the numbers of electron were not equal, the equations would need to be multiplied to obtain equal numbers of electrons gained and lost.

The emf is the electrical potential that can be calculated from the standard potentials.

$E^ o_{ c e l l} = E^ o_{ r e d u c t i o n} − E^ o_{ o x i d a t i o n} = 0.34 V − ( − 0.14 V ) = 0.48 V$