Ir E N equals o so V, what is E write the nee ionic equation for rlse reaction t

Question

Ir E N equals o so V, what is E write the nee ionic equation for rlse reaction that occurs in the cell studied by the studenti The student adds 6 M NH, to he Caso, solution umtil the cua. on is essentially all con verted to Cu(NHDa ion. The cell voltage, E increases to 0.92 v and the copper electrode is still negative. Find the concentration of Cu ion in the cell use Equation (10m. Show your calculations M. 8. In question 7 the ICu(NH s about 0.05 M and INH l is about 3 M. Given these values and ICu2 l from question 7. determine K for the following reaction. Show your calculations. Cu(NH 4 NH

Explanation / Answer

Cell potential = Reduction potential of the half cell at Cathode – Reduction potential of the half cell at Anode

E0cell = E0red, cathode - E0reduction, anode

From the standard reduction potential table, we know that E0red( Cu2+) = 0.34 V

E0red( Ag+) = 0.8 V

If Cu electrode was cathode and Ag electrode was anode,

E0cell = 0.34 – 0.8 = -0.46 V

But the given value of E0cell = 0.45 V

Hence, the Ag electrode is a cathode and Cu electrode is an anode.

In a galvanic cell, anode is considered negative while cathode is positive. Note, it is given that Cu is negative. Hence this cell is a galvanic cell and not electrolytic cell. The electrons flow from anode to cathode as oxidation takes place at anode while reduction takes place as cathode. At anode Cu metal which has 0 oxidation state is oxidised to give Cu2+ ion which has +2 oxidation state. At cathode, the Ag+ ion which has +1 oxidation state is reduced to elemental silver which has 0 oxidation state.

The half equations are given as:

Ag+(aq) + e– ------> Ag(s)………………………………..Reduction

Cu(s) ------> Cu2+ (aq) + 2e–……………….......Oxidation

Multiplying the reduction equation by 2 and adding the reduction and oxidation equations we get,

2Ag+(aq) + Cu(s) +2e-------> Cu2+(aq) + 2e- + 2Ag(s)

Overall reaction

2Ag+(aq) + Cu(s) ------> Cu2+(aq) + 2Ag(s)

Cell potential = Potential of the half cell at Cathode - Potential of the half cell at Anode

E0cell = E0cathode - E0anode

The voltmeter explicitly measures the overall redox potential between the two half-cells. The voltmeter has two probes, a red and black probe and records the voltage difference between the red probe and the black probe. If the measured voltage is 0.45 V and E0cathode is 0.00 V and A has 0.34 V, then E0anode will be -0.45 V.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.