Home
Class 12
CHEMISTRY
For the following electrochemical cell a...

For the following electrochemical cell at 298 k<
`Pt_(s) H_(g)(1bar)| H^(+)_(1M)||M _(aq)^(4+)`, `+(aq)^(2+) |Pt_(s)E _(cell) = 0.092V` When `([M_(aq)^(2+)])/([M_(aq)^(4+)] )=10^(X)`.
Given : `E_(M^(4+)//M^(2+))^(@) = 0.151 V`, `2.303(RT)/(F) = 0.059V ]`
The value of X is ,

A

`(-2)`

B

`(-1)`

C

1

D

2

Text Solution

AI Generated Solution

To solve the given electrochemical cell problem, we will use the Nernst equation and the information provided. Let's break it down step by step. ### Step 1: Understand the Cell Reaction The electrochemical cell consists of two half-reactions: 1. At the anode, hydrogen gas is oxidized: \[ \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \] ...
Promotional Banner

Similar Questions

Explore conceptually related problems

For the following electrochemical cell at 298 K, pt (s) |H_(2) (g, 1 " bar" )|H^(+) (aq.), M^(2+) (aq.)| Pt (s) E_("cell") = 0.92 V when ([M^(2+) (aq.)])/([M^(4+) (aq.)]) = 10^(x) Given : E_(M^(4+)//M^(2+))^(@) = 0.151 V, 2.303 (RT)/(F) = 0.059V The value of x is :

For the following electrochemical cell at 298K Pt(s)+H_(2)(g,1"bar") |H^(+) (aq,1M)||M^(4+)(aq),M^(2+)(aq)|Pt(s) E_(cell) = 0.092 V when ([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x) Given, E_(M^(4+)//M^(2+))^(@) = 0.151V, 2.303 (RT)/(F) = 0.059 The value of x is-

If electrode potential of following cell: Pt_(s)|Fe_(aq)^(2+),Fe_(aq)^(3+)||MnO_(4(aq))^(-),Mn_(aq)^(2+),H_(aq)^(+)|Pt_(s) is X then calculate value of 20X. [Given: E_(MnO_(4)^(-)|Mn^(2+) = 1.51 V, E_(Fe^(3+)|Fe^(2+))^(@) = 0.78 V, (2.303 RT)/F = 0.06]

Calculate the emf of the following cell at 298K : Fe(s)|Fe^(2+)(0.001M)||H^+(1M)|H_2(g)(1"bar"),Pt(s) (Given E_("Cell")^@=+0.44V)

The overall reaction electrochemical cell at 298K. Ag(s)|AgI(s)|I^(-)(aq)||Cl^(-)(aq)|Hg_(2)Cl_(2)|Hg(l)|Pt(s) [Given: E_(Cl^(-)|Hg_(2)Cl_(2)|Hg)^(@)=0.26V. E_(Ag^(+)|Ag)^(@)=0.8V . K_(sp)(Agl)=10^(-16) and (2.303RT)/(F)=0.06 ] E_("cell")^(@) of the above cell is: