Question

20) For the cell shown, the measured cell potential, ?cell, is −0.3541 V at 25 °C.

Pt(s) | H2(g,0.899 atm) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s)

The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, ?o, are

2H+(aq)+2e−⟶H2(g)?o=0.00 V

Cd2+(aq)+2e−⟶Cd(s)?o=−0.403 V

Calculate the H+ concentration.

Answer 1

H⁺ concentration = 0.141 M

Explanation

The overall reaction is : H₂ (g) + Cd²⁺ (aq) 2 H⁺ (aq) + Cd(s)

According to Nernst equation,

E_cell = E^o_cell - (0.0591 V / n) * log([H⁺]² / [pH₂][Cd²⁺])

-0.3541 V = -0.403 V - (0.0591 V / 2) * log([H⁺]² / (0.899 atm) * (1.00 M))

(0.0591 V / 2) * log([H⁺]² / (0.899 atm) * (1.00 M)) = -0.403 V + 0.3541 V

(0.0591 V / 2) * log([H⁺]² / (0.899 atm) * (1.00 M)) = -0.0489 V

log([H⁺]² / (0.899 atm) * (1.00 M)) = (-0.0489 V) * (2 / 0.0591 V)

log([H⁺]² / (0.899 atm) * (1.00 M)) = -1.65

[H⁺]² / (0.899 atm) * (1.00 M) = 10^-1.65

[H⁺]² / (0.899 atm) * (1.00 M) = 0.02214

[H⁺]² = (0.02214) * (0.899 atm) * (1.00 M)

[H⁺]² = 0.0199 M²

[H⁺] = (0.0199 M²)^1/2

[H⁺] = 0.141 M