Question

"The oxidation potential of Zn, Cu, Ag, H, and Ni
are 0.76,-0.34, -0.80, 0, 0.55 volt respectively.
Which of the following reaction will provide
maximum voltage ?
(1) Zn + Cu²+→ Cu + Zn2+
(2) Zn + 2Ag → 2Ag + Zn2+
(3) H2 + Cu2+ →2H+ + Cu
(4) H2 + Ni2+ →2H+ + Ni​

Answer 1

Hi there,

♤ Keys:

■ Anode at which the oxidation occures

■ Cathode at which the reduction occures

■ EMF of cell = E anode - E cathode by the oxidation potential

♤ Answer:

◇1- EMF of [Zn/Cu] cell = E (Zn) - E (Cu) = 0.76 - (-0.34) = + 1.1 v

◇2- EMF of [Zn/Ag] cell = E( Zn) - E (Ag) =

0.76 - (-0.80) = +1.56 v

◇3- EMF of [H/Cu] cell = E (H) - E (Cu) =

0.0 - (-0.34) = +0.34 v

◇ 4- EMF of [H/Ni] cell = E (H) - E (Ni) =

0.0 - ( 0.55) = -0.55 v

☆ So the answer is

Zn + 2Ag+ => Zn+2 + 2 Ag

♡ Hope the answer is helpful

♡ Plz mark it as brainliest

Answer 2

The reaction  $Zn + 2Ag^{+} \rightarrow 2Ag + Zn^{2+}$ will provide maximum voltage.

Explanation:

More is the value of E.M.F of a chemical reaction more will be voltage provided by it.

Mathematically,    E.M.F = $E_{cathode} - E_{anode}$

Reduction potential values are opposite in sign of oxidation potential.

(1)  For the reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$, reduction of copper is taking place and oxidation of zinc is taking place. Hence, emf of this reaction is as follows.

            E.M.F = $E_{cathode} - E_{anode}$ = 0.34 - (-0.76)

                      = 1.1 V

(2)  For the reaction $Zn + 2Ag^{+} \rightarrow 2Ag + Zn^{2+}$, reduction of silver is taking place and oxidation of zinc is taking place. Hence, emf of this reaction is as follows.

           E.M.F = $E_{cathode} - E_{anode}$ = 0.80 - (-0.76)

                      = 1.56 V

(3)  For the reaction $H_{2} + Cu^{2+} \rightarrow 2H^{+} + Cu$, reduction of copper is taking place and oxidation of hydrogen is taking place. Hence, emf of this reaction is as follows.

           E.M.F = $E_{cathode} - E_{anode}$ = 0.34 - (0)

                      = 0.34 V

(4)  For the reaction $H_{2} + Ni^{2+} \rightarrow 2H^{+} + Ni$, reduction of nickel is taking place and oxidation of hydrogen is taking place. Hence, emf of this reaction is as follows.

           E.M.F = $E_{cathode} - E_{anode}$ = 0.55 - (0)

                      = 0.55 V

Hence, emf value is maximum for reaction (2). Therefore, the reaction  $Zn + 2Ag^{+} \rightarrow 2Ag + Zn^{2+}$ will provide maximum voltage.

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