Represent the half cells or electrodes for the following reactions :(i

Represent the half cells or electrodes for the following reactions :(i) Zn2+(aq) + 2e- → Zn(s)(ii) \(Ag^+_{(aq)}\)+ e- → Ag(s)(iii) \(4H^+_{(aq)}\)+ 4e- → 2H2(g)(iv) \(Sn^{4+}_{(aq)}\)+ 2e- → \(Sn^{2+}_{(aq)}\) (v) \(\frac{1}{2}\)Cl2(g) + e- → Cl-(aq)

Answer»

Half cell reaction	Half cell (or electrode)
(i) Zn²⁺_(aq) + 2e^- → Zn_(s)	Zn²⁺_(aq)\| Zn_(s)
(ii) \(Ag^+_{(aq)}\)+ e^- → Ag_(s)	\(Ag^+_{(aq)}\)\| Ag_(s)
(iii) \(4H^+_{(aq)}\)+ 4e^- → 2H_2(g)	\(H^+_{(aq)}\)\|H₂(g,\(P_{H_2}\))\| Pt
(iv) \(Sn^{4+}_{(aq)}\)+ 2e^- → \(Sn^{2+}_{(aq)}\)	\(Sn^{2+}_{(aq)}\),\(Sn^{4+}_{(aq)}\)\| Pt
(v) \(\frac{1}{2}\)Cl_2(g) + e^- → Cl^-_(aq)	Cl^-_(aq)\|Cl₂(g,Pcl₂)\|Pt

Discussion

No Comment Found

Related InterviewSolutions

Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of a solution for a weak and a strong electrolyte.
Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:A. `Lambda_(m (NH_(4)Cl ))^(@) + Lambda_(m (NaCl))^(@) - Lambda_(m(NaOH))^(@)`B. `Lambda_(m (NaOH))^(@) + Lambda_(m(NaCl))^(@) - Lambda_(m(NH_(4)Cl))^(@)`C. `Lambda_(m(NH_(4)OH))^(@) + Lambda_(m (NH_(4)Cl))^(@) - Lambda_(m(HCl))^(@)`D. `Lambda_(m (NH_(4)Cl))^(@) + Lambda_(m (NaOH))^(@) - Lambda_(m(NaCl))^(@)`
Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:A. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(Na_(4)Cl)-Lambda_(m)^(@)(NaOH)`B. `Lambda_(m)^(@)(NaOH)+Lambda_(m)^(@)(NaCl)-Lambda_(m)^(@)(NH_(4)Cl)`C. `Lambda_(m)^(@)(NH_(4)OH)+Lambda_(m)^(@)(NH_(4)Cl)-Lambda_(m)^(@)(HCl)`D. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`
For the cell reaction `Cu^(2+)(C_(1),aq.)+Zn(s)hArrZn^(2+)(C_(2),aq)+Cu(s)` of an electrochemical cell, the change in free energy `(DeltaG)` of a given temperature is a function ofA. `lnC_(1)`B. `lnC_(2)//C_(1)`C. `ln (C_(1) + C_(2))`D. `ln C_(2)`
Use of lithium metal as an electrode in high energy density batteries is due to:A. Lithium is highest elementB. Lithium has highest oxidation potentialC. Lithium is quite reactiveD. Lithium does not corrode readily
The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follow : `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2), Delta_(r)G= +960 kJ mol^(-1)` The potential difference needed for the electrolytic reduction of aluminium oxide `(Al_(2)O_(3))` at `500^(@)C` isA. `5.0 V`B. `4.5 V`C. `3.0 V`D. `2.5 V`
Based on the data given below, the correct order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt A1`B. `Fe^(2+) lt Al lt Br^(-)`C. `Al^(-) lt Br^(-) lt Fe^(2+)`D. `Al^(-) lt Fe^(2+) lt Br^(-)`
Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^(-)`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^-`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
What are elementary reactions ?

Represent the half cells or electrodes for the following reactions :(i) Zn2+(aq) + 2e- → Zn(s)(ii) \(Ag^+_{(aq)}\)+ e- → Ag(s)(iii) \(4H^+_{(aq)}\)+ 4e- → 2H2(g)(iv) \(Sn^{4+}_{(aq)}\)+ 2e- → \(Sn^{2+}_{(aq)}\) (v) \(\frac{1}{2}\)Cl2(g) + e- → Cl-(aq)

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment