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Page No 73: - Chapter 3 Electrochemistry Intext Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


Question 3.4:

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Answer:

For hydrogen electrode, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6234/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_96797bd.gif , it is given that pH = 10

∴[H+] = 10−10 M

Now, using Nernst equation:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6234/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_56c137c3.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6234/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_m2be9aa1b.gif  https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6234/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_78ed9f0b.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6234/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_7be69ac0.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6234/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_m2bd4c672.gif

= −0.0591 log 1010

= −0.591 V

Question 3.5:

Calculate the emf of the cell in which the following reaction takes place:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6235/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_m40455851.gif

Given that https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6235/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_e96fdf6.gif = 1.05 V

Answer:

Applying Nernst equation we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6235/NCERT(INTEXT)_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

Question 3.6:

The cell in which the following reactions occurs:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6236/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_43a1beca.gif

has

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6236/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_1a3843ff.gif = 0.236 V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer:

Here, n = 2, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6236/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_36b4a02e.gif  T = 298 K

We know that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6236/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_m1674e026.gif

= −2 × 96487 × 0.236

= −45541.864 J mol−1

= −45.54 kJ mol−1

Again, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6236/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_m5b7431e5.gif −2.303RT log Kc

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6236/NCERT(INTEXT)_17-11-08_Utpal_12_Chemistry_3_15_html_m239a813b.gif

= 7.981

Kc = Antilog (7.981)

= 9.57 × 107

Also Read : Page-No-84:-Chapter-3-Electrochemistry-Intext-Solutions-class-12-ncert-solutions-Chemistry

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