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QUESTION 4 Using the table below and the Nearnst Equation E-E° (0.0592/n)logQ Find the voltage of the cell: Cu2+(aq) + Ni(s) ? Cu(s) + Ni2+(aq) at 25° C if the concentrations of the soluble species are [Cu2+1-0.050 M and [Ni]- 1.40 M Reduction Reaction Sn2++ 2e-? Sn Cu2+ + 2e- ?Cu Reduction Potential, Ecell 0.14 Volts 0.34 Volts 0.25 Volts +0.77 Volts Fe3+ + e-? Fe2+ 0.50 V 0.54 V 0.62 V 0.66 V

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Answer #1

Answer

0.54V

Explanation

Oxidation

Ni(s) - - - - - - > Ni2+(aq) + 2e E°red = - 0.25V

Reduction

Cu2+(aq) + 2e - - - - - > Cu(s) E°red= + 0.34V

overall

Cu2+(aq) + Ni(s) - - - - - > Cu(s) + Ni2+(aq)

E°cell = + 0.34V - (-0.25V)

E°cell = 0.59V

Q = [Ni2+]/[Cu2+] = 1.40M/0.050M = 28

n= no of electron transfer, 2

Nernst equation is

Ecell = E°cell - (0.0592V/n)logQ

Applying the values

Ecell = 0.59V - (0.0592V/2)log28

Ecell = 0.59V - 0.04V

Ecell = 0.55V

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