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 [tex]E^o_{cell} = 2.35 V[/tex]
Explanation:
From the question we are told that
The ionic equation is
[tex]3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}[/tex]
Now under standard conditions the reduction half reaction is
[tex]Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V[/tex]
And the oxidation half reaction is
[tex]Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V[/tex]
The emf of this cell under standard conditions is mathematically represented as
[tex]E^o_{cell} = E^o _r - E^o _o[/tex]
substituting values
[tex]E^o_{cell} = 1.61 - (- 0.74)[/tex]
[tex]E^o_{cell} = 2.35 V[/tex]
