Home
Class 12
CHEMISTRY
(10Cl^(-)(aq) + 2MnO(4)^(-)(aq) + 16H^(+...

`(10Cl^(-)(aq) + 2MnO_(4)^(-)(aq) + 16H^(+)(aq) rightarrow 5Cl_(2)(g) + 2Mn^(2+) + 8H_2O)(l)`
The value of `E^(@)` for this reaction at `25^(@)C` is 0.15 V. What is the value of K for this reaction?

Text Solution

Verified by Experts

`therefore-nFE^(@)= - 2.303RTlog K`
`thereforeK=-antilog[(nE^(@))/(0.059)]`
`K=antilog[(10xx0.15)/(0.059)] 2.4xx10^25` `thereforeKxx10^(-24)2.4 xx10^(25)xx10^(-24) 24`
Promotional Banner

Similar Questions

Explore conceptually related problems

10Cl^(-)(aq) + 2MnO_(4)^(-)(aq) + 16H^(+)(aq) rightarrow 5Cl_(2)(g) + 2Mn^(2+0()aq) + 8H_92)O(l) The value of E^(@) for this reaction at 25^(@)C is 0.15 V. What is the value of K for this reaction?

Complete the following: 2MnO_4^(-)(aq) + 5H_2O_2 (aq) to 2Mn^(2+) + 8H_2O+ …..

__ Sn^(2+) (aq)+__ MnO_(4)^(-) (aq)+__ H^(+) (aq) rarr __ Sn^(4+) (aq) +__ MnO^(2+) (aq) __ H_(2)O(l) When this equation for the reaction of Sn^(2+) (aq) and MnO_(4)^(-) (aq) is balanced correctly, what is the ratio, Sn^(2+)//MnO_(4)^(-) ?

Sn^(4+)(aq) + 2e^(-) rightarrow Sn^(2+)(aq) E^(@) = 0.15V Cr^(3+)(aq) + e^(-) rightarrow Cr^(2+)(aq) E^(@) = 0.41V According to the standard reduction potentials above, what is the value of E^(@) for the reaction below? 2Cr^(3+)(aq) + Sn^(2+)(aq) rightarrow 2Cr^(2+)(aq) + Sn^(4+)(aq)

The reaction H_(2)O_(2)(aq)+3I^(-)(aq)+2H^(+)(aq)rarrI_(3)^(-)(aq)+2H_(2)O(l) has a rate law : rate = k[H_(2)O_(2)][I^(-)] . What is the order of the reaction with respect to H^(+) , and what is the overall order of the reaction?

The standard potential for the reaction, Cl_(2)(g) + 2Br^(-)(aq) rightarrow Br_(2)(l) + 2Cl^(-)(aq) is 0.283 volts. What is the equilibrium consant for this reaction at 25^(@) C?

In the reaction ClO_(3)^(-)(aq)+5Cl^(-)(aq)+6H^(+)(aq)rarr3Cl_(2)(g)+3H_(2)O(l) the oxidizing and reducing agents are , respectively :