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CBSE Class 12-science Answered

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Asked by Balbir | 02 Oct, 2019, 12:29: AM
answered-by-expert Expert Answer
EMF of cell can be calculated by Nernst equation for the following reactions.
Ag left parenthesis NH subscript 3 right parenthesis subscript 2 to the power of plus space space space space end exponent plus straight e to the power of minus space left right double arrow space Ag space plus 2 NH subscript 3 After space applying space Nernst space equation minus 0.799 equals straight E to the power of straight o minus.0591 cross times log space straight K subscript ins left square bracket Ag left parenthesis NH subscript 3 right parenthesis subscript 2 to the power of plus space space end exponent right square bracket 0.799 equals straight E to the power of straight o minus.0591 cross times log space left parenthesis 6.02 cross times 10 to the power of negative 8 end exponent right parenthesis 0.799 space equals straight E to the power of straight o minus left parenthesis 0.799 minus 8 right parenthesis cross times 0.0591 space 0.799 equals straight E to the power of straight o minus left parenthesis negative 7.201 cross times 0.0591 right parenthesis straight E to the power of straight o space equals 0.799 minus 0.4255 space space space space space equals 0.372 straight V In space second space chemical space equation minus Ag left parenthesis CN right parenthesis subscript 2 to the power of minus space plus space straight e to the power of minus space left right double arrow space Ag plus space 2 CN to the power of minus After space applying space Nernst space equation minus 0.799 equals straight E to the power of straight o minus.0591 cross times log space straight K subscript ins left square bracket Ag left parenthesis CN right parenthesis subscript 2 to the power of minus right square bracket 0.799 equals straight E to the power of straight o minus space.0591 cross times log left parenthesis 1.99 cross times 10 to the power of negative 19 end exponent right parenthesis 0.799 equals straight E to the power of straight o minus.0591 cross times left square bracket log 1.995 minus 19 log 10 right square bracket straight E to the power of straight o equals 0.799 minus 1.10517 space space space equals negative 0.307 straight V
Answered by | 03 Oct, 2019, 02:47: PM

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CBSE 12-science - Chemistry
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Asked by yashwanthgowdakn4 | 22 Feb, 2024, 09:14: PM
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CBSE 12-science - Chemistry
Can you upload notes on a nernst equations explanation
Asked by keerthana.d.cst.2022 | 22 Aug, 2023, 08:18: PM
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CBSE 12-science - Chemistry
Q. 6. Calculate the emf of the following cell at 298 K Cr(s)/Cr³+ (0.1M)//Fe²+ (0.01M)/Fe(s) [Given: Eºcell = + 0.30 V]
Asked by amankumar2004112 | 04 May, 2022, 01:12: PM
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CBSE 12-science - Chemistry
Please answer this question.
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Asked by Balbir | 02 Oct, 2019, 12:29: AM
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CBSE 12-science - Chemistry
For what concentration of Ag+ will the EMF of te given cell be zero at 25 C .If the concentration of Cu2+ is 0.1 M. EoAg+ = 0.80v and Eo Cu2+ = 0.34 V
Asked by Jsiajaijiaia | 27 Aug, 2019, 12:23: AM
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CBSE 12-science - Chemistry
Please answer. 
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Asked by prakriti12oct | 14 Aug, 2019, 12:46: AM
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CBSE 12-science - Chemistry
Electrochemistry What's 'n' in the equation:           E°(cell) = 0.0591/n log Kc  
Asked by rakeshraghav33 | 19 May, 2018, 03:55: PM
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CBSE 12-science - Chemistry
Predict whether the following redox reaction is feasible under the standard conditions or not.              Sn2+(aq) + Cu(s) →   Sn(s) + Cu2+(aq)
Asked by Topperlearning User | 22 Jun, 2016, 02:42: PM
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CBSE 12-science - Chemistry
Zn(s) | Zn2+(0.01M) || Pb2+(1.OM)|Pb(s) Given: Eo Pb2+/Pb = -0.12V and Eo Zn2+/Zn = -0.76V, what is the cell potential of the cell?
Asked by Topperlearning User | 22 Jun, 2016, 02:41: PM
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CBSE 12-science - Chemistry
Calculate the reduction potential for the following half cell reaction at 298 K.             Ag+(aq) + e- →  Ag(s) Given that   (Ag+) = )0.1M and E° = + 0.80V
Asked by Topperlearning User | 22 Jun, 2016, 02:41: PM
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