You have a voltammetry cell in which the cathode is cu in a solution of 0.01 m cu²⁺. the anode is cd, in a solution of 0.01 m cd²⁺. the cell is producing a current of 0.100 a. the cell has a resistance of 4.00 ω. a.) what would be the electrochemical cell potential if no current was flowing? b.) what is the potential of the cell when you account for ir drop?

A) The emf of the cell when no current was flowing is 0.4 V

B) The potential of the cell when account for IR drop is < 0.4 V

Explanation:

Given,

The current of the cell, I = 0.100 A

The internal resistance of the cell, R = 4 Ω

The electrochemical cell potential is given by the formula,

                                                     V = IR

                                                         = 0.1 x 4

                                                         = 0.4 V

The electrochemical potential of the cell if no current was flowing is V = 0.4 volts

The IR drop occurs when a wire is connected across the terminals.

The potential of the cell after the IR drop is always < 0.4 V

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5