Calculate the emf of the Electrochemical Cell

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.
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