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1896 United States presidential election in Rhode Island

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1896 United States presidential election in Rhode Island

← 1892 November 3, 1896 1900 →
 
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Home state Ohio Nebraska
Running mate Garret Hobart Arthur Sewall
Electoral vote 4 0
Popular vote 37,437 14,459
Percentage 68.33% 26.39%

County Results
McKinley
  60-70%
  70-80%


President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

The 1896 United States presidential election in Rhode Island took place on November 3, 1896, as part of the 1896 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the Republican nominee, former governor of Ohio William McKinley, over the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan. McKinley won the state by a wide margin of 41.94%.

Bryan, running on a platform of free silver, appealed strongly to Western miners and farmers in the 1896 election, but had little appeal in Northeastern states like Rhode Island.

With 68.33% of the popular vote, Rhode Island would be McKinley's fourth strongest victory in terms of percentage in the popular vote after Vermont, neighboring Massachusetts and New Hampshire.[1]

Bryan would lose Rhode Island to McKinley again four years later and would later lose the state again in 1908 to William Howard Taft.

Results

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1896 United States presidential election in Rhode Island[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican William McKinley of Ohio Garret Hobart of New Jersey 37,437 68.33% 4 100.00%
Democratic William Jennings Bryan of Nebraska Arthur Sewall of Maine 14,459 26.39% 0 0.00%
National Democratic John McAuley Palmer of Illinois Simon Bolivar Buckner of Kentucky 1,166 2.13% 0 0.00%
Prohibition Joshua Levering of Maryland Hale Johnson of Illinois 1,160 2.12% 0 0.00%
Socialist Labor Charles Horatio Matchett of New York Matthew Maguire of New Jersey 558 1.02% 0 0.00%
N/A Others Others 5 0.01% 0 0.00%
Total 54,785 100.00% 4 100.00%

See also

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References

[edit]
  1. ^ "1896 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. ^ "1896 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.