Use the following half-reactions to construct a voltaic cell:
cr3+(aq) + 3e−\rightarrow→ cr(s) eo = −0.73 v
ag+(aq) + e−\rightarrow→ ag(s) eo = 0.80 v
determine the balanced overall redox reaction, and calculate eocell.
3 ag(s) + cr3+(aq) \rightarrow→ 3 ag+(aq) + cr(s), eocell = −1.53 v
3 ag+(aq) + cr(s) \rightarrow→ 3 ag(s) + cr3+(aq),eocell = +3.13 v
3 ag(s) + cr3+(aq) \rightarrow → 3 ag+(aq) + cr(s), eocell = −3.13 v
ag(s) + cr3+(aq) \rightarrow→ ag+(aq) + cr(s), eocell = −1.53 v
3 ag+(aq) + cr(s) \rightarrow→ 3 ag(s) + cr3+(aq), eocell = +1.53 v
ag+(aq) + cr(s) \rightarrow→ ag(s) + cr3+(aq), eocell = +1.53 v

The correct answer is

Explanation:

We are given:

The substance having highest positive potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction:

Reduction half reaction:       ( × 3)

Net equation:  

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

Hence, the correct answer is

you gotta add the following molecules so we can give u an answer

explanation: