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com mi.rds 1 Quiclet Apple Cloud Yahoo Bing Google Wikipedia Facebook Twitter Linkedin The Weather Channel Yolp California State Unive dino - CHEM 216-Summer16-MINK | Activities and Due Dates | Homework S o 9/6/2016 11:00 PM 0 14.2/209/3/2016 10:41 PM Print-Cal uator ba Pundic Table Gradebook a Sapling Learning Calculate the celt potential for the following reaction as written at 25.00 given tatlZn NP]- 00100 M. Standard reduction potentials can be found here. 0.874 M and Number E- 588 There is additional foodbac available View this teedback by cicking on te dvider bar bar again to hide the additional feedback Close Incorrect

Explanation / Answer

there is adirect equation

E (cell) = Eo (cell) - 0.0591 / n x log [Zn2+]/[Ni2+]

first find the Eo of the cell

Eo ( Zn2+/Zn) = -0.76 V

Eo (Ni2+ / Ni) = -0.25 V

Eo ( cell) = Eo ( cathode) - Eo(anode)

Eo ( cell) = -0.25 V - (-0.76 V)

Eo ( cell) = -0.25+0.76 = 0.51 V

n = no.of electrons transffered = 2

F = 96500 Coulombs

[Zn2+] = 0.874 M

[Ni2+] = 0.01 M

plug in all these values in above equation

E (cell) = 0.51 - 0.0591 / 2 x log[ 0.874 / 0.01]

E (cell) = 0.51 - 0.057

E (cell) = 0.453V

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