Question

What is the calculated value of the cell potential at 298 K for an electrochemical cell...

What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the H2 pressure is 5.00×10-3 bar, the H+ concentration is 1.46 M, and the Zn2+concentration is 3.42×10-4 M ?

2H+(aq) + Zn(s) ----> H2(g) + Zn2+(aq)

The cell reaction as written above is product-favored (spontaneous) for the concentrations given: _______(true/false)?

i)

0.780V

Explanation

2H+(aq) + Zn(s) ------> H2(g) + Zn2+(aq)

Reactant quotient, Q = PH2 × [Zn2+]/[H+]2

Q = 0.005bar × 0+.000342M/(1.46)2

Q = 8.022×10-7

Oxidation half reaction (at anode)

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

Reduction half reaction (at cathode)

2H+(aq) + 2e --------> H2(g) E°red = 0.000V

cell = E°red,cathode - E°red,anode = 0.000V - (- 0.762V) = 0.762V

Number of electron transfer ,n = 2

at 25°C , Nernst equation is as follows

Ecell = E°cell - (0.0592V/n)logQ

Ecell = 0.762V - (0.0592V/2)log(8.022×10-7)

Ecell = 0.762V + 0.18V

Ecell = 0.780V

ii)

True

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