What is the emf of the electrochemical cell Y(s) | Y2+(1.235) || X3+(7.627) |X(s) based on the given standard reduction potentials?
The emf of the electrochemical cell Y(s) | Y2+(1.235) || X3+(7.627) |X(s) is -0.822 V. It can be calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode.
Explanation:
Electrochemical Cell Components:
An electrochemical cell consists of two half-cells, namely a cathode and an anode. In this case, the given electrochemical cell is composed of the following components:
- Cathode: Y(s) | Y2+(1.235) with a reduction potential of -0.812 V
- Anode: X3+(7.627) | X(s) with a reduction potential of 0.01 V
Calculation of Emf:
The emf of the cell can be calculated using the Nernst equation, which is given by: Ecell = Eo(cell) - (0.0592/n)log(Q), where Ecell is the emf of the cell, Eo(cell) is the standard reduction potential of the cell, n is the number of electrons transferred, and Q is the reaction quotient.
In this scenario, the reduction potential of the cathode (Y) is -0.812 V, and the reduction potential of the anode (X) is 0.01 V. By subtracting the anode potential from the cathode potential, we get:
Emf = E(cathode) - E(anode) = -0.812 V - 0.01 V = -0.822 V
Therefore, the emf of the electrochemical cell Y(s) | Y2+(1.235) || X3+(7.627) | X(s) is calculated to be -0.822 V.