Question

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)

Answer: _____ V ?

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

Answer 1

i)

0.780V

Explanation

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

Reactant quotient, Q = P_H2 × [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

E°_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

E_cell = E°_cell - (0.0592V/n)logQ

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

E_cell = 0.762V + 0.18V

E_cell = 0.780V

ii)

True