Quiz 1 on Electrochemistry

Given the standard reduction potentials, E^o of iron is -0.44V and E^o of oxygen is 1.23V.

what will be the emf of the cell?

• -0.79V

• -1.67V

• 1.67V

• 0.79V

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Galvanic cell converts

• Chemical energy into electrical energy

• Electrical energy into chemical energy

• a metal from its elemental form to combined form

• Electrolyte into individual ions

At equilibrium

• Cell potential will be zero

• Equilibrium constant becomes equal to electrode potential

• Equilibrium constant becomes zero

• Cell potential becomes unity

For this electrochemical cell, what is the expected observation at cathode?

• Copper metal dissolves

• reddish brown solid deposited

• Silvery grey solid deposited

• Silver metal dissolves

What is the direction of electron flow.

• From Zn to Cu through the external circuit

• From Zn to Cu through salt bridge

• From Cu to Zn through the external circuit

• From Cu to Zn through salt bridge

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As an element is oxidized, its oxidation number

• decreases as electrons are lost

• increases as electrons are lost

• decreases as electrons are gained

• increases as electrons are gained

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NGSS.HS-PS1-1

NGSS.HS-PS1-2

Electrons are gained at the

• anode

• cathode

Electrons always flow from

• cathode to anode

• anode to cathode

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What is an anode?

• It is the electrode where oxidation takes place.

• It is the electrode where reduction takes place.

• It is an aqueous solution containing an electrode.

• It is the salt bridge that connects half-cells.

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NGSS.HS-PS2-5

Nernst equation FOR (i) Mg(s) | Mg2+(0.001M) || Al3+(0.0001 M) | Al(s)

• E_CELL=E⁰_CELL-0.0591/6 log[Mg2+]³/[Al3+]²

• E_CELL=E⁰_CELL-0.0591/6 log[Al3+]²/[Mg2+]³

• E_CELL=E⁰_CELL+0.0591/3 log[Mg2+]³/[Al3+]²

• E_CELL=E⁰_CELL-0.0591/3 log[Mg2+] /[Al3+]

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NGSS.HS-PS1-4

NGSS.HS-PS3-1