Question

[61 Consider the following concentration cell consisting of aluminum electrodes in Al(NO3)3 solutions. (a) Label the anode and cathode and show the direction that electrons will travel on the diagram. (b) Determine the initial voltage read by the volt meter. (c) Determine the concentration of A3t in each cell when the volt meter reads zero [AM] =

Answer 1

Solution: A concentration cell  is an electrolytic cell having two half-cells with the same electrodes, but differing in concentrations. It acts to dilute the more concentrated solution and concentrate the more dilute solution.This is achieved by transferring the electrons from the half cell with the lower concentration to the half cell with the higher concentration.

A) In concentration cell, oxidation occurs at left hand electrode ( anode) and reduction on right hand electrode (cathode). Thus, left side eletrode is anode and right side electrode is cathode. Since, in the given concentration cell of Aluminium, the left hand electrode has lower concentration (0.015M) of Al⁺³ and higher concentration on right hand electrode (0.495M).Hence, electrons are traveling from left hand electrode to right hand electrode.

B) The initial voltage read by the Volameter is calculated by Nernst equation, since (n=3 for Al)

Ecell = (RT/nF) ln [ C2/C1] = (0.0591/3)x log [0.495/0.015] = 0.299 V

C) In this type of reaction, there are two electrodes which are involved. These are known as the anode and the cathode, or the left and right side, respectively. The anode is the side which is losing electrons (oxidation) while the cathode is the side which is gaining electrons (reduction).

Ecell = (RT/nF) ln [ C2/C1]

when Ecell = 0, then C1 = C2

Then the concentrations of Al³⁺ in both the half cells are equal.