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

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.