Election to the United States House of Representatives 1796

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In the election to the United States House of Representatives in 1796, the House of Representatives was elected on various election days in the United States from August 12, 1796 . The election was part of the general election for the 5th United States Congress that year, in which a third of the US Senators were elected. At the same time the presidential election of 1796 took place, which the federalist John Adams won.

At the time of the election, the United States consisted of 16 states . The number of MPs to be elected was 106. The distribution of seats in the House of Representatives was based on the 1790 census .

Women and slaves were neither eligible nor eligible to vote. Free African Americans were also excluded from voting in many states . The right to vote for free men was also tied to a certain property or tax revenue.

The election resulted in a clear victory for the federalists, who now took over the majority in the House of Representatives with 57 seats.

Election result

Total : 106

Distribution of seats
  
A total of 106 seats
  • Federalist Party : 57
  • Democratic Republican Party : 49

The results of the last regular elections of 1794 are in brackets. Changes during the legislative period that do not affect the elections are not included in these figures, but are included in the article on the 5th Congress in the section on the members of the House of Representatives noted the corresponding names of the MPs. The same applies to elections in states that did not join the Union until after the beginning of the legislative period. As a result, the sources sometimes contain different information, as changes during the legislative period were sometimes incorporated into the figures and sometimes not.

See also

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