Chemistry, 30.08.2019 00:30 mixcolin0002

it is expected that a chemical reaction will occur when copper metal is combined with aqueous zinc sulfate. explain why there will be no reaction when zinc metal and aqueous copper sulfate solution are combined. identify the anode and the cathode, assuming a voltaic cell is constructed. note: be careful in the calculation of the standard cell potential ( eo cathode - eo anode). do not change the sign of the given reduction potential. the sign is already taken care of using the formula for calculating the standard cell potential.
standard reduction potential: cu2+(aq) + 2e- > cu(s) eo = -0.34 v
zn2+(aq) + 2e- > zn(s) eo = -0.76 v
standard cell potential = eo cathode - eo anode
although , combination of the reactants do not result in voltaic cells, it can be predicted if the reaction is spontaneous, based on the standard reduction potential.

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Answer from: hiplikedyani

$E_{cell}$ = +ve, reaction is spontaneous

$E_{cell}$ = -ve, reaction is non spontaneous

$E_{cell}$ = 0, reaction is in equilibrium

Case 1: when copper metal is combined with aqueous zinc sulfate.

$Cu+ZnSO_4\rightarrow Zn+CuSO_4$

Here copper is undergoing oxidation ad thus acts as anode and zinc is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both E^0 are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Cu^{2+}/Cu]}$

E^0=-0.76-(+0.34)=-1.10V

Thus as $E_{cell}$ is negative , the reaction is non spontaneous.

Case 2: when zinc metal and aqueous copper sulfate solution are combined.

$Zn+CuSO_4\rightarrow Cu+ZnSO_4$

Here zinc is undergoing oxidation ad thus acts as anode and copper is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both E^0 are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Zn^{2+}/Zn]}$

E^0=+0.34-(-0.76)=+1.10V

Thus as $E_{cell}$ is positive , the reaction is spontaneous.

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

the answer is c.) 1575 j.

explanation:

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