For the following galvanic cell, represented in line notation, determine what balanced half-reactions occur at each electrode. (Use the lowest possible whole number coefficients. Include states-of-matter under the given conditions in your answer.) Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)anode half-reaction:cathode half-reaction:

Respuesta :

Answer:

Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

Explanation:

Let's consider the notation of a galvanic cell.

Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)

On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).

The half-reactions are:

Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.

Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)

ACCESS MORE