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The reaction Zn^(2+) + 2e^(–)rarrZn(s) h...

The reaction `Zn^(2+) + 2e^(–)rarrZn(s)` has a standard potential of `– 0.76 V`. This means:

A

Zn can't replace hydrogen from acids

B

Zn is a reducing agent

C

Zn is an oxidising agent

D

`Zn^(2+)` ions are a reducing agent

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The correct Answer is:
B
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Standard reduction electrode potential of Zn^(2+)//Zn is -0.76V . This means:

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Calculate the emf of the cell. Zn∣ Zn^(2+)(0.001M)∣∣Cu^(2+) (0.1M)∣Cu .The standard potential of Cu/ Cu^(2+) half-cell is +0.34 and Zn/ Zn^(2+) is -0.76 V.

The electrode potential, E^(@) , for the reduction of MnO_(4)^(-)" to "Mn^(2+) in acidic medium is +1.51V . Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for Zn^(2+), Ag^(+) and Au^(+) are given as Zn_((aq))^(2+)+2e^(-)rarrZn_((s)),E^(@)=-0.762V Ag_((aq))^(+)+e^(-)hArr Ag_((x)), E^(@)=+0.80V Au_((aq))^(+)+e^(-)hArr Au_((s)), E^(@)=+1.69V

For I_(2)+2e^(-) rarr 2I^(-) , standard reduction potential =+0.54 V. For 2Br^(-) rarr Br_(2)+2e^(-) , standard oxidation potential =-1.09V. For Fe rarr Fe^(2+)+2e^(-) , standard oxidation potential =+0.44V . Which of the following reactions is ( are ) spontaneous ?

(a) Calculate Delta G^@ for the reaction Zn (s) + Cu^(2+) (aq) rarr Zn^(2+) + Cu(s) Given: E^@ for (Zn^(2+))//Zn = -0.76 V and E^@ for (Cu^(2+))//Cu = +0.34 V R= 8.314 JK^-1 mol^(-1) F= 96500 Cmol^(-1) (b) give two advantages of fuel cells.

Calculate the voltage of following electrochemical cell under standard conditions. Name anode and cathode in each cell. (i) Zn - Ag (ii) Co - Ni. The standard reduction potentials of various half cells are: Zn^(2+)//Zn =-0.76V, Ag^(+)//Ag= +0.80V Co^(2+)//Co = -0.28V, Ni^(2+)//Ni= -0.25V

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