Cell reaction is spontaneous when
Three cell A, B and C has equilibrium constant in the ratio 1:4 : 9 respectively. Arrange the following cells in the order of increasing Gibbs free energy.
The correct answer is option A
The cells will be arranged as such 9 : 4: 1.
Explanation:
The equilibrium constant that is used here is represented with k.
The smaller the value of k the higher will be the value of energy according to Gibbs.
So, 1 has the highest Gibbs free energy while 9 has the least.
This is the energy of a chemical reaction.
It is represented by G.
Gibbs free energy change for a cell reaction is positive what does it indicates?
The correct answer is Option C.
Reactions with a negative ∆G release energy, which means that they can proceed without an energy input (are spontaneous). In contrast, reactions with a positive ∆G need an input of energy in order to take place (are nonspontaneous). Reactions with a positive ∆H and negative ∆S are nonspontaneous at all temperatures.
At equilibrium:
The correct answer is option A
E cell is 0 in equilibrium, that is Ecathode becomes equal to E anode ……….
EZero cell becomes zero when both the electrodes are of the same metal but of different concentration i.e for concentration cell……
We know that EZero cell is  Ezerocathode E zeroanode (since cathode and anode are same ) so EZero =0
In the equation, ΔG° = – nF E° cell ; F is:
The correct answer is Option B.
The relationship between ΔG^{o} and E^{o} is given by the following equation: ΔG^{o}=−nFE^{o}. Here, n is the number of moles of electrons and F is the Faraday constant.
Consider the cell reaction:
Cd(s)  Cd^{2+} (1.0 M)  Cu^{2+} (1.0 m)  Cu (s)
If we wish to make a cell with more positive voltage using the same substances, we should:
The correct answer is Option D.
Redox reaction:
Cd(s)→Cd^{2+}+2e
Cu^{2+}+2e→Cu(s)
E_{cell} = E°_{cell} − (0.059/2) log ([Cd^{2+}]/ [Cu^{2+}])
Decreases [Cd^{2+}] to 0.1M and increases [Cu^{2+}] to 1.0M
. The electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by:
The correct answer is Option B
Nernst equation is,
where, Q is the reaction quotient of the reaction
As, [M (s) ]=1
We get,
. For an equation: Ni(s) + 2Ag^{+}(aq) → Ni^{2+} (aq) + 2Ag(s) the Nernst equation is written as:
The correct answer is Option A.
Oxidation:
Ni_{(s)} > Ni^{2+}_{(aq)} + 2e^{}
Reduction:
2Ni^{+}_{(aq)} + 2e^{} > 2Aq_{(s)}
2e are in the above reaction so;
n = 2
We know that
E_{cell }= E^{o}_{cell} – (RT/nF) lnK_{c}
= E^{o}_{cell} – (RT/nF) ln ([Ni]^{2+} / [Ag^{+}]^{2}
Nernst equation for an electrode is based on the variation of electrode potential of an electrode with:
Nernst equation for an electrode is based on the variation of electrode potential of an electrode with temperature and concentration of electrolyte.
The free energy change for the following cell reaction is given as :
2Au^{3+} (aq) + 3Cu (s) → 2Au (s) + 3Cu^{2+} (aq)
The correct answer is Option D.
E^{O} = E^{O}_{Ca}^{2+}/ _{Ca}  E^{O}_{Au}^{2+}/ _{Au}
= 2.87  (1.50)
= 2.87  1.50
= 4.37 V
rG^{O} = nFE^{O}
= 6 FE^{O}




