Question

In: Chemistry

Given the following reaction, Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) E° = 1.10 V. Use...

Given the following reaction, Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) E° = 1.10 V.

Use the Nernst equation to calculate the cell potential for the cell described with standard line notation below.

Zn|Zn2+(0.5082 M)||Cu2+(0.2699 M)|Cu

Units are not required. Report answer to three decimal places.

Please explain all of your steps! Thanks!

Solutions

Expert Solution

Oxidation Half cell

Zn(s) --------> Zn2+(aq) + 2e Anode E°red= -0.762V

Reduction Half cell

Cu2+(aq) + 2e --------> Cu(s) Cathode E° red = 0.337V

overall reaction is

Cu2+(aq) + Zn(s) -------> Cu(s) + Zn2+ (aq)

E°cell = E°red(Cathode) - E°red(Anode)

= 0.337V - (-0.762V)

= 1.10V

Nernst equation is

Ecell = E°cell - (RT/nF)lnQ

where,

R= 8.314(J/mol K )

T = 298K ( standard temperature )

number of electron transfer , n =2

Faraday constant, F = 96485C/mol

So, at standard temperature Nernst equation is

Ecell = E°cell - (0.0592V/n)logQ

Q = [ Zn2+ ] / [ Cu2+ ]

= 0.5082M/0.2699M

= 1.8829

So,

Ecell = 1.10V - (0.0592/2) log 1.8829

= 1.10 V - 0.0081V

= 1.092V

So, the cell potential is 1.092V

queen_honey_blossom answered 1 year ago

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