# 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