Please provide a step by step solution and answers for the 3 Voltages answer req

Question

Please provide a step by step solution and answers for the 3 Voltages answer required. NOTE MY ANSWER ARE INCORRECT, please show me how to correct. thank you.

consider the cell: Cu

nsider the cell: CulCu2 (0.00571 M)lPb2 (0.00603 M)IPb Calculate the half cell reduction potential at 298 K at the cathode. (Refer to your textbook for standard reduction potentials.) Number .6896 Calculate the half cell reduction potential at 298 K at the anode. Number -.345 What is the initial potential needed to provide a current of 0.0660 A, if the resistance of the cell is 4.36 ? (Assume T-298 Number .288

Explanation / Answer

Solution:

Given data,

Cu|Cu2+(0.00571 M) ||Pb2+(0.00603 M) | Pb        -------------------- (A)

From standard potential table we get,

Pb2+ + 2e- à Pb                - 0.13 V          (anode)

Cu2+ + 2e- à Cu                 + 0.34 V          (cathode)

The equation to calculate overall cell potential is,

Ecell = (Ecathode – Eanode) + (0.0592 V/n) log Keq         --------------------- (1)

The unknowns in the equation (1) are calculated from (A),

Keq = [0.00603]/[0.00571] = 1.05604

Log Keq = log (1.05604) = 0.2368

n = 2 as redox is 2e-

by substituting the above values in equation (1) we get,

Ecell = (0.34 – (-0.13)) +(0.0592/2) (0.02368

Ecell = 0.47 V

From ohm’s law we know V = I*R     { I = 0.0660 A and R = 4.36 ohm}

V = 0.0660 * 4.36 = 0.305 V

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