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Goryan [66]
3 years ago
6

The magnetic field of a bar magnet Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an

answer. a Always points from north to south, whether we look outside the magnet or inside the magnet. b Always points from south to north, whether we look outside the magnet or inside the magnet. c Points from north to south outside the magnet, but south to north inside the magnet. d Points from south to north outside the magnet, but north to south inside the magnet.
Physics
1 answer:
viva [34]3 years ago
7 0

Answer:

A: Always points from north to south, whether we look outside the magnet or inside the magnet

Explanation:

When we freely suspend a bar magnet, it's magnetic field will always rest in the north to south direction due to the fact that the Earth's magnetic south pole usually lies in the direction of the geographic north while the Earth's magnetic north pole usually lies in the direction of the geographic south.

Now, we all know that in magnets, like poles will repel and unlike poles will attract. What this now means is that the magnet will always point from the north to the south direction.

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xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi
slega [8]

Answer:

Approximately \rm 90\; kJ.

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (E^{\circ}(\text{cell})) is equal to

E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode}).

There are two half-reactions in this question. \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu and \rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of E^{\circ}(\text{cell}) should be positive.

In this case, E^{\circ}(\text{cell}) is positive only if \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu is the reaction takes place at the cathode. The net reaction would be

\rm Cu^{2+} + Pb \to Cu + Pb^{2+}.

Its cell potential would be equal to 0.339 - (-0.130) = \rm 0.469\; V.

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell}),

where

  • n is the number moles of electrons transferred for each mole of the reaction. In this case the value of n is 2 as in the half-reactions.
  • F is Faraday's Constant (approximately 96485.33212\; \rm C \cdot mol^{-1}.)

\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}.

5 0
3 years ago
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