Question

A voltaic cell is composed of Ni/Ni²⁺ and Co/Co²⁺ half-cells. The initial concentration of [Ni²⁺] is 1.00 M, and [Co²⁺] = 0.25 M. The half-cell potentials are as follows:

Co⁺(aq)+2e⁻→Co(s)E°=−0.28V
Ni⁺(aq)+2e⁻→Ni(s)E°=−0.25V
Ni⁺(aq)+Co(s)→Ni(s)+Co⁺(aq)

Calculate the initial Ecell.

Answer 1

The initial Ecell can be calculated using the formula Ecell = E°cathode - E°anode. In this case, the E°cathode is -0.28V and the E°anode is -(-0.25V) = 0.25V. Therefore, the initial Ecell is -0.28V - 0.25V = -0.53V.

Step-by-step explanation:

The initial Ecell can be calculated using the formula Ecell = E°cathode - E°anode. In this case, the E°cathode is -0.28V and the E°anode is -(-0.25V) = 0.25V. Therefore, the initial Ecell is -0.28V - 0.25V = -0.53V.

To calculate the initial Ecell of a voltaic cell composed of Ni/Ni²⁺ and Co/Co²⁺ half-cells with given half-cell potentials, you can use the Nernst equation. However, since initial concentrations are already at standard conditions (1 M for Ni²⁺) and no non-standard conditions were mentioned, standard potential values can be directly used to find the Ecell. Remember that the half-cell with the higher reduction potential will act as the cathode. The given potentials are for the reductions of Co and Ni:

Co⁺⁴(aq) + 2e⁻ → Co(s), E° = -0.28 V (Anode)

Ni²⁺(aq) + 2e⁻ → Ni(s), E° = -0.25 V (Cathode)

Therefore, we reverse the sign for the oxidation (anode) and keep it as is for the reduction (cathode). Now we can calculate the initial Ecell as:

Ecell = E°cathode - E°anode

Ecell = (-0.25 V) - (-0.28 V) = 0.03 V

So, the initial Ecell for the voltaic cell is 0.03 V.