The following half-reactions are used in the zinc-air battery. Zn(OH) ₄^2- (aq) + 2 e- → Zn(s) + 4 OH-(aq) E° = -1.199 V
O₂(g) + 2 H₂O(l) + 4 e- → 4 OH- (aq) E° = +0.401 V
How much charge is transferred per gram Zn(s) in this voltaic cell?

Question

Quizplus · Accepted Answer

The charge transferred per mole of Zn can be calculated by using the Faraday constant, which is 96,485 C/mol e-. First, balance the reactions and reverse the first reaction to get the oxidation half-reaction: Zn(s) + 4 OH-(aq) → Zn(OH)4^(2-)(aq) + 2 e- E° = +1.199 V 4 O₂(g) + 2 H₂O(l) + 8 e- → 8 OH- (aq) E° = -0.401 V Now, we can use the Nernst equation to calculate the cell potential at standard conditions: Ecell = E°red + E°ox = (+0.401 V) - (+1.199 V) = -0.798 V The amount of charge transferred per mole of Zn can be calculated using the formula: Q = n * F * Ecell where n is the number of electrons transferred in the balanced reaction and F is the Faraday constant. Since 2 electrons are transferred for every Zn atom, n = 2. Now, we need to find the number of moles of Zn that will produce 1 gram of Zn. The molar mass of Zn is 65.38 g/mol, so 1 g of Zn is equivalent to 1/65.38 mol of Zn. Putting it all together: Q = n * F * Ecell * (1/65.38) mol Q = 2 * 96,485 C/mol * (-0.798 V) * (1/65.38) mol Q = -2,950 C Q (in absolute value) = 2,950 C = 2.95 × 10^3 C Therefore, the answer is B, 2.95 × 10^3 C or 2.95 × 10^13 µC.

The Following Half-Reactions Are Used in the Zinc-Air Battery