Question

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer 1

Complete Question

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer:

The value is $E^o_{cell} = 2.35 V$

Explanation:

From the question we are told that

   The ionic equation is  

               $3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}$

Now under standard conditions the reduction  half reaction  is

      $Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V$

And the oxidation half reaction is

      $Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V$

The emf of this cell under standard conditions  is mathematically represented as

     $E^o_{cell} = E^o _r - E^o _o$

substituting values

     $E^o_{cell} = 1.61 - (- 0.74)$

    $E^o_{cell} = 2.35 V$