Home
Class 11
CHEMISTRY
What will be the spontaneous reaction be...

What will be the spontaneous reaction between the following half cell reactions ?
(i) `Cr^(3+)(aq)+3e^(-) rarr Cr(s) " "E^@=-0.74V`
(ii) `MnO_2(s)+4H^(+)+2e^(-)rarr Mn^(2+)(aq)+2H_2O(l)" "E^@=1.28V`
Calculate `E_("cell")^@`

Text Solution

Verified by Experts

Since the reduction potential reaction (ii) is more than that of reaction (i), reaction (ii) will occur as reduction . Therefore , reaction (i) should be written as oxidation . To obtain the net reaction , we multiply by appropriate coefficients so that electrons get cancelled .
`MnO_2(s)+4H^(+)(aq)_+2e^(-)rarrMn^(2+)(aq)+2H_2O(l)]xx3`
`" "Cr(s)rarrCr^(3+)(aq)+3e^(-)]xx2`
`2Cr(s)+3MnO_2(s)+12H^(+)(aq)rarr2Cr^(3+)(aq)+3Mn^(2+)(aq)+6H_2O`
`E_("cell")^@=E_("Substance reduced")-E_("Substance oxidised")`
`=1.28-(-0.74)=2.02V`
Promotional Banner

Topper's Solved these Questions

  • REDOX REACTIONS

    MODERN PUBLICATION|Exercise Practice problems|17 Videos

  • REDOX REACTIONS

    MODERN PUBLICATION|Exercise Conceptual Questions|20 Videos

  • ORGANIC CHEMISTRY: BASIC PRINCIPLES AND TECHNIQUES

    MODERN PUBLICATION|Exercise (Competition File) MULTIPLE CHOICE QUESTIONS ( based on the given. passage/comprehension )|8 Videos

  • S-BLOCK ELEMENTS ( ALKALI AND ALKALINE EARTH METALS )

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|13 Videos

Similar Questions

Explore conceptually related problems

What will be the spontaneous reaction when the following half reactions are combined ? (i) Fe^(3+) + e^(-) to Fe^(2+) , (ii) MnO_(4)^(-) + 8 H^(+) + 5 e^(-) to Mn^(2+) + 4 H_(2) O , (E^(Theta) = + 1.49 V ) What is the value of E_(cell)^(Theta) ?

If the half cell reactions are given as (i) Fe^(2+)(aq)+2e^(-)rarrFe(s),E^(@)=0.44V (ii) 2H^(+)(aq)+ 1//2O_(2)(g)+2e^(-)rarrH_(2)O(l),E^(@)=+1.23V The E^(@) for the reaction.

Read the following passage for the evaluation of E^(@) when different number of electrons are involved. Consider addition of the following half reactions (1) Fe^(3+)(aq)+3e^(-)rarr Fe(s) E_(1)^(@)=0.45V (2) Fe(s)rarr Fe^(2+)(aq)+2e^(-) E_(2)^(@)=-0.04V (3) Fe^(3+)(aq)+e^(-)rarr Fe^(2+)(aq) E_(3)^(@)= ? Because half-reactions (1) and (2) contains a different number of electrons, the net reaction (3) is another half-reaction and E_(3)^(@) can't be obtained simply by adding E_(1)^(@) and E_(3)^(@) . The free - energy changes however, are additive because G is a state function : Delta G_(3)^(@)=Delta G_(1)^(@)+Delta G_(2)^(@) For the reactions (1) MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2)+4H_(2)O E^(@)=1.51 V (2) MnO_(2)+4H^(+)+2e^(-)rarr Mn^(2+)2H_(2)O E^(@)=1.32 then for the reaction (3) MnO_(4)^(-)+4H^(+)+3e^(-)rarrMnO_(2)+2H_(2)O, E^(@) is

The standard reduction potentials at 298K for the following half cells are given : ZN^(2+)(aq)+ 2e^(-)Zn(s) , E^(@) = - 0.762V Cr^(3+)(aq)+ 3e^(-) Cr(s) , E^(@) = - 0.740V 2H^(+) (aq)+2e^(-) H_(2)(g) , E^(@)= 0.000V Fe^(3+)(aq)+ e^(-) Fe^(2+)(aq) , E^(@) = 0.770V which is the strongest reducing agent ?

The standard electrode potential for the following reaction is +1.33V . What is the potential at pH =2.0 ? Cr_(2)O_(7)^(2-)(aq,1M)+14H^(+)(aq)+6e^(-) rarr 2Cr^(3+)(aq. 1M)+7H_(2)O(l)

Consider the following half reactions : PbO_(2)(s)+4H^(o+)(aq)+SO_(4)^(2-)+2e^(-)rarrPbSO_(4)9s)+2H_(2)O" "E^(c-)=+1.70V PbSO_(4)(s)+2e^(-)rarr Pb(s)+SO_(4)^(2-)(aq)" "E^(c-)=-0.31V a. Calculate the value of E^(c-) for the cell. b. Calculate the voltage generated by the cell if [H^(o+)]=0.10M and [SO_(4)^(2-)]=2.0M c. What voltage is generated by the cell when it is at chemical equilibrium ?

Given below are the half -cell reactions Mn^(2+) + 2e^(-) rarr Mn, E^@ =- 1. 18 V 2 (Mn^(3+) = E^(-) rarr Mn^(2+)). E^@ = + 1.5 V The E^@ for Mn^(2+) rarr Mn+ 2 Mn^(3+) will be.