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Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
For hydrogen electrode, _17-11-08_Utpal_12_Chemistry_3_15_html_96797bd.gif){kind=small}
, it is given that pH = 10
∴[H+] = 10−10 M
Now, using Nernst equation:
_17-11-08_Utpal_12_Chemistry_3_15_html_56c137c3.gif){kind=small}
= _17-11-08_Utpal_12_Chemistry_3_15_html_m2be9aa1b.gif){kind=small}
_17-11-08_Utpal_12_Chemistry_3_15_html_78ed9f0b.gif){kind=small}
_17-11-08_Utpal_12_Chemistry_3_15_html_7be69ac0.gif){kind=small}
_17-11-08_Utpal_12_Chemistry_3_15_html_m2bd4c672.gif){kind=small}
= −0.0591 log 1010
= −0.591 V
Calculate the emf of the cell in which the following reaction takes place:
_17-11-08_Utpal_12_Chemistry_3_15_html_m40455851.gif){kind=small}
Given that _17-11-08_Utpal_12_Chemistry_3_15_html_e96fdf6.gif){kind=small}
= 1.05 V
Applying Nernst equation we have:
_17-11-08_Utpal_12_Chemistry_3_15_html_4ad268a1.gif)
= 1.05 − 0.02955 log 4 × 104
= 1.05 − 0.02955 (log 10000 + log 4)
= 1.05 − 0.02955 (4 + 0.6021)
= 0.914 V
The cell in which the following reactions occurs:
_17-11-08_Utpal_12_Chemistry_3_15_html_43a1beca.gif){kind=small}
has _17-11-08_Utpal_12_Chemistry_3_15_html_1a3843ff.gif){kind=small}
= 0.236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Here, n = 2, _17-11-08_Utpal_12_Chemistry_3_15_html_36b4a02e.gif){kind=small}
T = 298 K
We know that:
_17-11-08_Utpal_12_Chemistry_3_15_html_m1674e026.gif){kind=small}
= −2 × 96487 × 0.236
= −45541.864 J mol−1
= −45.54 kJ mol−1
Again, _17-11-08_Utpal_12_Chemistry_3_15_html_m5b7431e5.gif){kind=small}
−2.303RT log Kc
_17-11-08_Utpal_12_Chemistry_3_15_html_m239a813b.gif)
= 7.981
∴Kc = Antilog (7.981)
= 9.57 × 107
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