Question

a voltaic cell is set up at 25 degree Celsius with the following half cell Ag+ (. 001 mole) and Cu2+ (.01 mole) what will be the voltage of this cell? Ecell=0.46 voly

Answer 1

Hey !!

QUESTION

A voltaic cell is set up at 25 degree Celsius with the following half cell Ag⁺ (0.10 mole) and Cu2⁺ (0.10 M) Cu .What will be the voltage of this cell ? [ E⁰cell = 0.46 V, log 10⁵ = 5 ]

ANSWER

  Cu + 2Ag⁺ -----> Cu⁺ + 2Ag

Half cell reactions

Cathode (reduction)

2Ag⁺ (0.001 M) + 2e⁻ ----> 2Ag(s)

Cu(s) ----> Cu²⁺ (0.10 M) + 2e⁻

                       n = 2

        E⁰cell = 0.46 V

       Ecell = E⁰cell - 0.059 / n log [Cu²⁺] / [Ag⁺]²

    = 0.46 - $\frac{0.059}{2} log\frac{0.01}{0.001}$²

     = 0.46 - $\frac{0.059}{2} log$10⁵

     = 0.046 - 0.059 / 2 × 5

      = 0.3125 V

Good luck !!